What is the vertices (x^2)/(27)-(y^2)/(216)=1 ?

The vertices (x^2)/(27)-(y^2)/(216)=1 is (3sqrt(3),0),(-3sqrt(3),0)

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vertices (x^2)/(27)-(y^2)/(216)=1

vertices

\frac{x ^{2}}{27} - \frac{y ^{2}}{216} = 1

(33, 0), (- 33, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

\frac{x ^{2}}{27} - \frac{y ^{2}}{216} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 33, b = 66

(0 + 33, 0), (0 - 33, 0)

Refine

(33, 0), (- 33, 0)

\frac{y ^{2}}{25} - x^{2} = 1

- 9 x^{2} + 16 y^{2} - 36 x - 160 y + 220 = 0

\frac{y ^{2}}{4} - \frac{x ^{2}}{21} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{81} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{36} = 1

What is the vertices (x^2)/(27)-(y^2)/(216)=1 ?
The vertices (x^2)/(27)-(y^2)/(216)=1 is (3sqrt(3),0),(-3sqrt(3),0)